Alright guys (and girls?)

I'm designing an audio spectrum analyser. It'll be a very crude design, with a function no more important than displaying it's output on a set of LEDs to represent low frequency to high frequency so that they "dance" in response to music. Not a new idea I know, and yes I know there'll already be circuits out there for this but I'm not after that I want to do the design work myself.

So anyway I'm currently looking into some 2nd order filters to see if the pass band will be narrow enough for what I want. I've got a very mathematical book in front of me and it's describing the operation of a 2nd order butterworth filter in terms of the following transfer function;

F(S) = Ho/Bn(S), where Ho is the d.c. gain and Bn(S) is the polynomial for the nth order filter.

I don't care where the equation comes from or how it's derived, I only care how I can use it. My question is, what does the (S) in brackets represent?

AudioGuru??

Brian

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Electronics Test Development Engineer
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When you study a filter in the frequency domain, you usually use the Laplace transform (of its impulse response). 's' ia the complex variable in the Laplace domain.

The polynomial in the denominator can be factored and your transfer function F(s) can be expressed as follows:
F(s) = H0 / [ k (s-p1) (s-p2) ... (s-pn) ]
where p1,p2,...,pn are the poles of your filter.

 

I guessed it was to do with laplace because I remember doing something with that before. At the time I never managed to understand laplace properly, think I'm going to have to revisit it. Doh!

Brian

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Electronics Test Development Engineer
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Tables!...and partial fractions. Bleh. Or matlab.

 

Using two Butterworth filters (a lowpass and a highpass) uses a lot of parts to make a poor bandpass filter. Make bandpass filters from a single opamp instead. A few Multiple Feedback Bandpass Filters are used in what you are doing and in EQ circuits.

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Uncle $crooge